But that doesn't stop us from speculating about their existence.

The photon is aloof from its peers. Massless, it races through the Universe at the speed of light, un-aging from the point of view of our slow-moving selves. Being massless imparts immortality of a sort. Photons are the ultimate in stable particles—they can't spontaneously decay into lighter particle, because they are the lightest particle around.

Fellow conspiracy theorists, prepare your tin-foil hats. What if all that weren't true? What if the photon had mass, did age, and could decay? Could it be possible? According to German physicist Julian Heeck, the answer is "Yes." But that "yes" is limited by some seriously accurate experimental data.

Having given away the punch line, let's go back and explain the joke (I'm told that jokes are always funnier when you explain them). We begin with a bunch of equations. The nice thing about equations is that you can always bolt extra bits on them. This is essentially what you do when you consider the photon having mass. But when you do that, things go very wrong. Suddenly, infinities produced by the equations that used to go away hang around, drink all your beer, and insist on watching re-runs of Murder, She Wrote. It just doesn't bear thinking about.

The story would end there if there were only one way to add bits to a mathematical model. However, there are ways of creating well-behaved photons that have some mass. If the photon had a serious amount of mass, though, our Universe would be a very different place. Indeed, some early calculations showed that the mass of a photon had to be smaller than 10-54kg (for reference, the electron is 10-31kg).

Having mass has other consequences, chief among them that the photon could decay to some other less massive particle—a neutrino, for instance. The question that Heeck asked (from my experimental physicist's point of view) was "How would you know if a photon decayed?"

Relativity plays an enormous role here. Photons, even if they are massive, are travelling very fast. For them, time runs very slow. Even if they had a rather short natural lifetime, they experience time so slowly that we would never observe them to decay in the laboratory. Even light from stars that are hundreds or thousands of light years away probably wouldn't have a single photon decay in the entire time that they have been emitting photons. No, if we are going to determine the lifetime of the photon, it will have to be from the oldest source of light in the Universe: the cosmic microwave background (CMB).

CMB photons come from the moment when the Universe cooled enough to become transparent. If the CMB photons have mass, though, the photons will have a range of energies, which also means that they travel at slightly different speeds and consequently decay at slightly different rates. This favors the survival of high-energy photons compared to low-energy photons.

In the paper, this was the subject of feverish calculation, resulting in a graph that shows what the CMB spectrum would look like for different photon decay rates. And this is where Heeck was rather clever. He compared his calculations to the spectrum obtained from the Cosmic Background Explorer (COBE) data. The COBE data is one of the most accurate data sets ever obtained, meaning that there is very little wiggle room in any fit to the data.

Surprisingly, the COBE data doesn't do a good job of telling us anything about the photon mass. Because of the shape of the spectrum, the COBE data does tell us that if the photon does have mass, then its lifetime is on the order of a few years. That is, if we were to hold a bunch of photons still relative to us, half of them would have decayed into neutrinos after a few years. But once you take into account the speed at which photons travel, those few years become a decay time of 1018 years for visible light (the age of the universe is 109 years).

The title of the paper asked the question "How stable is the photon?" I was disappointed that Heeck did not end the paper with "in conclusion, the photon is pretty stable." Nevertheless, he managed a few moments of humor. The first sentence of the abstract reads, "Yes, the photon" (quite necessary, as it turns out that I wondered if it was supposed to be "proton"). And anyone who writes "Still, unmeasurable small SM [standard model] rates never stopped anyone from looking for a signal..." knows the dark side of experimental physicists far too well.

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Chris Lee
Chris writes for Ars Technica's science section. A physicist by day and science writer by night, he specializes in quantum physics and optics. He is delocalised, living and working in Eindhoven and Enschede, the Netherlands. Emailchris.lee@arstechnica.com//Twitter@exMamaku

who had the unfortunate distinction of making a large number of important theoretical advances 20 to 30 years before these advances were made by the people who got the credit for them. It's possible that he was not taken seriously because he had the habit of marching around in the uniform of an Austrian cavalry officer.

Okay, it's been a long time...but I thought that special relativity-wise time dilation was something like 1/(sqrt(1-v^2/c^2)). If a photon is traveling at the speed of light (due to it being, well, a photon), that would seem to imply a rather infinite time dilation, none of this instantaneous 10^18 years stuff.

Similar problems with infinities seem to arise with mass-energy of a massive photon at the speed of light.

Clearly I'm missing something here, not surprising since I'm definitely not a physicist.

Okay, it's been a long time...but I thought that special relativity-wise time dilation was something like 1/(sqrt(1-v^2/c^2)). If a photon is traveling at the speed of light (due to it being, well, a photon), that would seem to imply a rather infinite time dilation, none of this instantaneous 10^18 years stuff.

Similar problems with infinities seem to arise with mass-energy of a massive photon at the speed of light.

Clearly I'm missing something here, not surprising since I'm definitely not a physicist.

My best guess is this: Photons would only move at c if they were massless. If they have an incredibly slight mass (which is needed for them to decay), then they'll move at almost-but-not-quite c, which removes the need to get a square root of negative 1.

Okay, it's been a long time...but I thought that special relativity-wise time dilation was something like 1/(sqrt(1-v^2/c^2)). If a photon is traveling at the speed of light (due to it being, well, a photon), that would seem to imply a rather infinite time dilation, none of this instantaneous 10^18 years stuff.

Similar problems with infinities seem to arise with mass-energy of a massive photon at the speed of light.

Clearly I'm missing something here, not surprising since I'm definitely not a physicist.

If photons were massive, they would not be traveling at the "speed of light" (ie., c), but some (minutely) lower speed. Yes, this would mean that the "speed of light" would not actually be the speed of light (particles); that's a nomenclature issue, not a definitional one.

I'm wondering, what if you could slow down the vacuum velocity of a single photon. But then again, it would cease to exist, since there's nothing it could decay into.

But, if the photon has a mass, we'd have to find the "true" yard stick of particle physics. The particle, that would travel at the maximum velocity, and therefore is created and destroyed simultaneously (from the photon point of view).

Interesting though. I think, I have to don my thinking hat to ponder about this.

So, in the end he was able to bracket the mass of the photon but not declare it massless? The fact that the half-life of a photon may be 10^9 times longer than the age of the universe so far is irrelevant. That only means we haven't seen many decay so far is all. But the implications of massive photons would be huge.

If they're massive how do you add more mass to them to force them to behave like the low-energy photons of the early CMB? I suspect that accomplishment would earn a Nobel.

I'm wondering, what if you could slow down the vacuum velocity of a single photon. But then again, it would cease to exist, since there's nothing it could decay into.

But, if the photon has a mass, we'd have to find the "true" yard stick of particle physics. The particle, that would travel at the maximum velocity, and therefore is created and destroyed simultaneously (from the photon point of view).

Interesting though. I think, I have to don my thinking hat to ponder about this.

I'm not too worried about it, I think we need more evidence and experimental verification to grant this theory traction.

...But therein lies the next big problem: if light decays, what does it decay into? Will it be a lepton, or something else?

Also, as I am not a physicist, I have a question: how can a photon have varying amounts of mass? I ask because I thought a single subatomic particle had a constant amount of mass.

Well, if anything moves at a speed greater than zero, its mass changes by the Lorentz factor. So if a photon moves at the speed of light (and has non-zero mass), its mass is theoretically infinite. However, if it moves at anything less than that, its mass is going to depend on the speed at which it's moving.

Okay, it's been a long time...but I thought that special relativity-wise time dilation was something like 1/(sqrt(1-v^2/c^2)). If a photon is traveling at the speed of light (due to it being, well, a photon), that would seem to imply a rather infinite time dilation, none of this instantaneous 10^18 years stuff.

Similar problems with infinities seem to arise with mass-energy of a massive photon at the speed of light.

Clearly I'm missing something here, not surprising since I'm definitely not a physicist.

Don't forget that the Lorentz factor you quote is for relative velocity. Additionally, since c is for speed of light in a vacuum and space is not homogenously an aboslute vacuum, v !== c.

Lastly, the Lorentzian time dilation you reference is itwself a simplification and does not account for gravitation, which also has a measurable effect.

So, no infinite time dilation involved, even if the photon is assumed to be massless.

...those few years become a decay time of 10^18 years for visible light (the age of the universe is 10^9 years).

This introduces more interesting conversations. For example, what was the universe up to during the illusory 14 x 10^9 years that we previously thought had elapsed since it was "born"? Most of the laggardly sources I am aware of cite the current "best estimate" of the age of the Universe as very close to 1.38 x 10^10 years. The Universe and I are (literally) not-so-suddenly missing something, here.

My initial thought is that wouldn't a massive photon react dramatically differently around a black hole? Heck, it might even dramatically change their formation. But whether and at what energy a photon can escape seems like it would be changed substantially if light particles didn't travel at the speed of light. (Having accepted that no physical circles have a d/circumference of pi, I don't have much problem with slower than light light.)

Okay, it's been a long time...but I thought that special relativity-wise time dilation was something like 1/(sqrt(1-v^2/c^2)). If a photon is traveling at the speed of light (due to it being, well, a photon), that would seem to imply a rather infinite time dilation, none of this instantaneous 10^18 years stuff.

Similar problems with infinities seem to arise with mass-energy of a massive photon at the speed of light.

Yes, that's why physicists will sometimes say that photons don't experience time at all (i.e. they don't age). From the point of view of the photon, the universe passes by all at once.

Of course, massless photons can't have a reference frame attached to them at all (in part because time doesn't pass for them), so using the Lorentz equations on them at all is... well, wrong.

Massive photons wouldn't travel at the "speed of light" at all, just very very very close to it.

samkass wrote:

My initial thought is that wouldn't a massive photon react dramatically differently around a black hole? Heck, it might even dramatically change their formation. But whether and at what energy a photon can escape seems like it would be changed substantially if light particles didn't travel at the speed of light.

The difference in speed between c (aka the "speed of light") and what massive photons would travel at would an insignificant difference in how photons interact with black holes. The event horizon, for example, would still be exactly where it is now, except the actual "place where light can escape" would be ever so slightly outside the event horizon. That... could introduce some issues, but not that we would have noticed yet. Very little light is produced that close to the horizon and escapes anyways (it may introduce some problems for Hawking radiation, but that's still theoretical anyways).

Don't forget that the Lorentz factor you quote is for relative velocity. Additionally, since c is for speed of light in a vacuum and space is not homogenously an aboslute vacuum, v !== c.

Light-waves propagate at different speeds in different media (phase velocity), but the photons still always move at c (the front velocity).

Quote:

Lastly, the Lorentzian time dilation you reference is itwself a simplification and does not account for gravitation, which also has a measurable effect.

Well, gravity is solely attractive these days, so any effect would merely work to slow time down even more.

Quote:

So, no infinite time dilation involved, even if the photon is assumed to be massless.

It's pretty well accepted that massless photons do not move along the time dimension. I'm not a physicist either, but Brain Greene is: "light does not get old; a photon that emerged from the big bang is the same age today as it was then. There is no passage of time at light speed" The Elegant Universe

they might be light, but there's a lot of them. Here's a thought: What if the constantly increasing expansion of the Universe is partly caused by these photons emitted by every star and pushing pressure on every other star and galaxies too far away to be attracted by the emitter's gravity?

they might be light, but there's a lot of them. Here's a thought: What if the constantly increasing expansion of the Universe is partly caused by these photons emitted by every star and pushing pressure on every other star and galaxies too far away to be attracted by the emitter's gravity?

The solid angle subtended by each star would fall off as the square of the distance between them - the exact same falloff one would see for a gravitational attraction. So, gravitational attraction and momentum exchange through photons would scale exactly the same.

There is something I definitely don't get about the concept of a photon with mass. I thought the entire point of special relativity was that the velocity of light is constant in all inertial reference frames, and so there is no way to determine a universal reference/rest frame. That requires massless photons that don't experience time.

But if photons have mass, then all of that is no longer true. Their velocity could NOT be constant in all inertial reference frames, and special relativity would not be true. You could determine an absolute rest frame.

Furthermore, and even nastier, since this would destroy a symmetry, by Noether's theorem it also destroys a conservation law, whatever is associated with losing the indistinguishability of inertial reference frames.

There is something I definitely don't get about the concept of a photon with mass. I thought the entire point of special relativity was that the velocity of light is constant in all inertial reference frames, and so there is no way to determine a universal reference/rest frame.

But if photons have mass, then all of that is no longer true. Their velocity could NOT be constant in all inertial reference frames, and special relativity would not be true. You could determine an absolute rest frame.

Furthermore, and even nastier, since this would destroy a symmetry, by Noether's theorem it also destroys a conservation law, whatever is associated with losing the indistinguishability of inertial reference frames.

Though it was likely cast as the speed of light being the same in all directions, it really means that massless particles travel at the same speed in all directions. That is neutrinos, gravity waves, EM waves, etc. all propagate through a perfect vacuum at c. However, IF photons have a finite (and damned small!) rest mass then, necessarily they will no longer be in that club. And the speed of light will be a misnomer. However, the difference between their actual speed and c will be so indistinguishably small that we won't be measuring it for a long time - unless as someone pointed out before that some black hole effect (or some other phenomena) scales the difference for us.

What would massive photons do to the overall mass of the Universe? Would it shift us from open/heat-death to closed?

No, not really. Photons actually already contribute to the mass density of the universe, even while themselves massless: a system of photons can have a rest mass, even when though the individual photons themselves do not, although cosmologists usually use energy density rather than mass density anyways (same thing really, just a factor of c^2 to convert).

The only difference it would make is when the photons start decaying, but since photons make up an insignificant part of the energy in the universe anyways, the effect would also be insignificant. Dark matter and dark energy influence the long-term future of the universe.

And no, decayed light can't serve as dark energy/dark matter (although it is possible it contributes a teensy-tiny immeasurably little bit of dark matter when it decays to neutrinos).

One thing I always struggle with... light is energy... and energy is mass (from E=mc^2). Where does the mass of the energy go if light is massless?

E=mc^2 isn't quite the whole story; there's another term that while normally negligible represents all of the energy in a photon. The expanded formula is:

E^2 = (mc^2)^2 + (pc)^2

where the last term represents the momentum of the system. At non-relativistic velocities for ordinary objects the amount of energy in their momentum is negligible compared to the energy in their rest mass (~20 megatons per kilogram). At relativistic velocities for massive objects you need to consider both terms, for massless particles like photons only the second term on the right remains and the equation simplifies to:

E = pc

where p (or the Greek letter rho if you're feeling pedantic and want to mess with unicode input) represents momentum.

No, not really. Photons actually already contribute to the mass density of the universe, even while themselves massless: a system of photons can have a rest mass, even when though the individual photons themselves do not, although cosmologists usually use energy density rather than mass density anyways (same thing really, just a factor of c^2 to convert).

The only difference it would make is when the photons start decaying, but since photons make up an insignificant part of the energy in the universe anyways, the effect would also be insignificant. Dark matter and dark energy influence the long-term future of the universe.

And no, decayed light can't serve as dark energy/dark matter (although it is possible it contributes a teensy-tiny immeasurably little bit of dark matter when it decays to neutrinos).

Thank you for the answer! I'm way outside my depth here - but I have a follow up, how does this involve dark matter? Aren't photons well in the normal matter group (and by normal I mean the stuff we see and interact with on a day to day basis). I thought dark matter was something else entirely.

I've always wondered how a black hole could capture photons as no matter how much the gravity is, something without mass wouldn't be influenced by it.But if photons do have mass it suddenly makes sense.

One thing I always struggle with... light is energy... and energy is mass (from E=mc^2). Where does the mass of the energy go if light is massless?

E=mc^2 isn't quite the whole story; there's another term that while normally negligible represents all of the energy in a photon. The expanded formula is:

E^2 = (mc^2)^2 + (pc)^2

Now now, don't take a shortcut and cheat the poster out of meaningful insight into the nature and definition of mass. Those two equations are not the same; the first one is not the same as the second but "incomplete" with a term hacked off just because it "normally" is pretty small.

E=mc^2 is the whole story for relativistic mass. It isn't an approximate, "normally" correct equation. It's exactly right, always. It is a fairly deep insight -- the apparent mass of a system is directly dependent on its energy, and therefore like energy will change depending on what reference frame you are measuring from.

Physicists don't usually use this definition of mass for two reasons:1) Like energy, it varies depending on who is doing the observing, so questions like "what is the mass of a proton" don't have singular answers.2) It is exactly the same as energy and thus redundant.

So the second equation, E^2 = (m0*c^2)^2 + (pc)^2 uses a different definition of mass (that I'm labeling m0), the invariant mass, which is the mass of the system relative to a frame that is not moving, and then separates out the contribution of mass/energy that is due to relative motion. All observers will agree the system has at least that much mass.

Now here's why it's cheating to just refer to the second equation as the "whole story" and leave the first to the historical dustbin of redundant concepts: Invariant mass includes the idea that motion is just another form of energy and therefore mass. As baloroth aluded to, in a system with multiple photons, there can be a center of momentum frame where the total momentum of the system is 0. In that case, the system would have a non-zero invariant mass which is precisely equal to the total E of the two photons / c^2.

So a box lined with a perfect mirror with photons bouncing back and forth would have a higher invariant mass, and weigh more on a scale, due to the energy of those photons. It would have a bigger m0, but a p still exactly equal to 0.

But when all you talk about is the "full" relativistic energy equation, while skipping the fuller discussion of what the invariant mass in that equation means, you make it sound like mass and momentum are entirely separate concepts and never the two shall meet. This is untrue. It's only relative momentum of the system as a whole that is wholly separate from the concept of invariant mass. The only time it is true that momentum and mass are completely separate concepts is in a system of exactly one fundamental particle.

I've always wondered how a black hole could capture photons as no matter how much the gravity is, something without mass wouldn't be influenced by it.But if photons do have mass it suddenly makes sense.

Well in the framework of General Relativity gravity is just a warping of space-time, and the photon (and all other objects under the influence of gravity) are traveling in "straight" lines, where "straight" isn't intuitive because the geometry is necessarily non-Euclidean.

Another way of looking at it is that in GR gravity is the consequence not of (invariant) mass but of energy. Photons have energy, and are therefore affected by gravity. They also effect gravity, but the energy density of photons in the universe is very tiny.

There's no reason to posit a rest mass of photons to have gravitational interaction with light make sense.

they can't spontaneously decay into lighter particle, because they are the lightest particle around.

That doesn't sounds logical enough. When photon is the lightest, it means photon has mass or it wouldn't be saying it is the lightest, right? When a particle has mass, no matter how light it is, there always something else is lighter.